1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Scalar field

  1. Sep 29, 2013 #1
    1. The problem statement, all variables and given/known data

    This is not really a problem but I was going over my lecture notes and I see [itex] \mathscr{H}=\frac{1}{2}\left(\pi^{2} + \vec{\nabla}\phi \cdot \vec{\nabla}\phi + m^{2}\phi^{2}\right) [/itex] and [itex]\frac{\partial\mathscr{H}}{\partial\phi} = -\nabla^{2}\phi + m^{2}\phi [/itex]

    2. Relevant equations



    3. The attempt at a solution

    I would think that [itex]\frac{\partial\mathscr{H}}{\partial\phi} = \nabla^{2}\phi + m^{2}\phi [/itex]. But I don't know where the minus sign is coming from.
     
  2. jcsd
  3. Sep 29, 2013 #2
    I just found a vector identity [itex] \vec{\nabla}\phi\cdot\vec{\nabla}\phi = \vec{\nabla}\cdot\left(\phi\vec{\nabla}\phi\right) - \phi\vec{\nabla}^{2}\phi [/itex]. I now see how the result follow.

    EDIT: I'm confused again. will the phi derivative of the first term vanish?
     
  4. Sep 30, 2013 #3
    If by "the first term" you mean the [itex]\pi^2[/itex] term, then yes, the [itex]\phi[/itex] derivative will kill that term; reason being [itex]\phi[/itex] doesn't appear in that term, only its time derivative.

    Edit: Sorry, I understand what you meant by "first term" now. The first term vanishes at infinity so you can ignore it.
     
    Last edited: Sep 30, 2013
  5. Sep 30, 2013 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    The 3-divergence in the Hamiltonian density (i.e. the first term in the RHS of the equality in post #2) can be discarded, since by integration of full space gives 0, because scalar fields are normally taken from the Schwartz space of test functions.
     
  6. Oct 2, 2013 #5


    That makes sense. I also realized that it can be integrated by parts using the same reasoning of vanishing at the boundaries. Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Scalar field
  1. Complex Scalar Field (Replies: 0)

  2. Energy of Scalar Field (Replies: 3)

Loading...